M O D U L E – I

S W I T C H I N G F U N C T I O N

Digital Design

Amit Kumar

Assistant Professor

SCSE,

Galgotias University, Greater Noida

Outline

• SWITCHING FUNCTION• CANONICAL AND STANDARD FORM• POS AND SOP FORM

A Boolean function consists of an algebraic expression formed with binary variables, the constants 0 and 1, the logic operation symbols, parenthesis, and an equal sign.

Example:

F(X,Y,Z) = X + Y’ Z or F = X + Y’ Z

X, Y and Z are Boolean variables A literal: The appearance of a variable or its complement in a

Boolean expression

A Boolean function can be represented with a truth table

A Boolean function can be represented with a logic circuit diagram composed of logic gates.

Boolean Functions

Switching algebra: Boolean algebra with the set of elements K = {0, 1}

If there are n variables, we can define switching functions.

Sixteen functions of two variables

A switching function can be represented by a table as above, or by a switching expression

as follows:

f0(A,B)= 0, f6(A,B) = AB' + A'B, f11(A,B) = AB + A'B + A'B' = A' + B, ...

Value of a function can be obtained by plugging in the values of all variables:

The value of f6 when A = 1 and B = 0 is: = 0 + 1 = 1.

AB f0 f1 f2 f3 f4 f5 f6 f7 f8 f9 f10 f11 f12 f13 f14 f15

00 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1

01 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1

10 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1

11 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1

22n

1 0 1 0 ' '

Switching Functions

Truth Tables

Shows the value of a function for all possible input combinations.

Truth tables for OR, AND, and NOT

ab f(a,b)=a+b ab f(a,b)=ab a f(a)=a'

00 0 00 0 0 1

01 1 01 0 1 0

10 1 10 0

11 1 11 1

Truth Tables

Truth tables for f(A,B,C) = AB + A'C + AC'

ABC f(A,B,C) ABC f(A,B,C)

000 0 FFF F

001 1 FFT T

010 0 FTF F

011 1 FTT T

100 1 TFF T

101 0 TFT F

110 1 TTF T

111 1 TTT T

Canonical Forms

Minterms and Maxterms Sum-of-Minterm (SOM) Canonical Form Product-of-Maxterm (POM) Canonical

Form Representation of Complements of

Functions Conversions between Representations

Minterms

Minterms are AND terms with every variable present in either true or complemented form.

Given that each binary variable may appear normal (e.g., x) or complemented (e.g., ), there are 2n minterms for n variables.

Example: Two variables (X and Y) produce2 x 2 = 4 combinations: (both normal) (X normal, Y complemented) (X complemented, Y normal) (both complemented)

Thus there are four minterms of two variables.

YX

XY

YX

YX

x

Maxterms

Maxterms are OR terms with every variable in true or complemented form.

Given that each binary variable may appear normal (e.g., x) or complemented (e.g., x), there are 2n maxterms for n variables.

Example: Two variables (X and Y) produce2 x 2 = 4 combinations: (both normal) (x normal, y complemented) (x complemented, y normal) (both complemented)

YX +

YX +

YX +

YX +

Two variable minterms and maxterms.

The minterm mi should evaluate to 1 for each combination of x and y.

The maxterm is the complement of the minterm

Minterms & Maxterms for 2 variables

x y Index Minterm Maxterm

0 0 0 m0 = x y M0 = x + y

0 1 1 m1 = x y M1 = x + y

1 0 2 m2 = x y M2 = x + y

1 1 3 m3 = x y M3 = x + y

Minterms & Maxterms for 3 variables

M3 = x + y + zm3 = x y z3110

M4 = x + y + zm4 = x y z4001

M5 = x + y + zm5 = x y z5101

M6 = x + y + zm6 = x y z6011

1

1

0

0

y

1

0

0

0

x

1

0

1

0

z

M7 = x + y + zm7 = x y z7

M2 = x + y + zm2 = x y z2

M1 = x + y + zm1 = x y z1

M0 = x + y + zm0 = x y z0

MaxtermMintermIndex

Maxterm Mi is the complement of minterm mi

Mi = mi and mi = Mi

Purpose of the Index

Minterms and Maxterms are designated with an index The index number corresponds to a binary pattern The index for the minterm or maxterm, expressed as a

binary number, is used to determine whether the variable is shown in the true or complemented form

For Minterms:• ‘1’ means the variable is “Not Complemented” and • ‘0’ means the variable is “Complemented”.

For Maxterms:• ‘0’ means the variable is “Not Complemented” and • ‘1’ means the variable is “Complemented”.

Standard Order All variables should be present in a minterm or

maxterm and should be listed in the same order (usually alphabetically)

Example: For variables a, b, c:• Maxterms (a + b + c), (a + b + c) are in standard order

• However, (b + a + c) is NOT in standard order

(a + c) does NOT contain all variables

• Minterms (a b c) and (a b c) are in standard order

• However, (b a c) is not in standard order

(a c) does not contain all variables

Sum-Of-Minterm (SOM) Sum-Of-Minterm (SOM) canonical form:

Sum of minterms of entries that evaluate to ‘1’

x y z F Minterm0 0 0 0

0 0 1 1 m1 = x y z

0 1 0 0

0 1 1 0

1 0 0 0

1 0 1 0

1 1 0 1 m6 = x y z

1 1 1 1 m7 = x y z

F = m1 + m6 + m7 = ∑ (1, 6, 7) = x y z + x y z + x y z

Focus on the ‘1’

entries

+ a b c d

Sum-Of-Minterm Examples

F(a, b, c, d) = ∑(2, 3, 6, 10, 11)

F(a, b, c, d) = m2 + m3 + m6 + m10 + m11

G(a, b, c, d) = ∑(0, 1, 12, 15)

G(a, b, c, d) = m0 + m1 + m12 + m15

+ a b c da b c d + a b c d+ a b c d + a b c d

+ a b c da b c d + a b c d

Product-Of-Maxterm (POM) Product-Of-Maxterm (POM) canonical form:

Product of maxterms of entries that evaluate to ‘0’

x y z F Maxterm0 0 0 1

0 0 1 1

0 1 0 0 M2 = (x + y + z)

0 1 1 1

1 0 0 0 M4 = (x + y + z)

1 0 1 1

1 1 0 0 M6 = (x + y + z)

1 1 1 1

Focus on the ‘0’

entries

F = M2·M4·M6 = ∏ (2, 4, 6) = (x+y+z) (x+y+z) (x+y+z)

F(a, b, c, d) = ∏(1, 3, 6, 11)

F(a, b, c, d) = M1 · M3 · M6 · M11

G(a, b, c, d) = ∏(0, 4, 12, 15)

G(a, b, c, d) = M0 · M4 · M12 · M15

Product-Of-Maxterm Examples

(a+b+c+d) (a+b+c+d) (a+b+c+d) (a+b+c+d)

(a+b+c+d) (a+b+c+d) (a+b+c+d) (a+b+c+d)

Observations

We can implement any function by "ORing" the minterms corresponding to the ‘1’ entries in the function table. A minterm evaluates to ‘1’ for its corresponding entry.

We can implement any function by "ANDing" the maxterms corresponding to ‘0’ entries in the function table. A maxterm evaluates to ‘0’ for its corresponding entry.

The same Boolean function can be expressed in two canonical ways: Sum-of-Minterms (SOM) and Product-of-Maxterms (POM).

If a Boolean function has fewer ‘1’ entries then the SOM canonical form will contain fewer literals than POM. However, if it has fewer ‘0’ entries then the POM form will have fewer literals than SOM.

Converting to Sum-of-Minterms Form

A function that is not in the Sum-of-Minterms form can be converted to that form by means of a truth table

Consider F = y + x z

x y z F Minterm0 0 0 1 m0 = x y z

0 0 1 1 m1 = x y z

0 1 0 1 m2 = x y z

0 1 1 0

1 0 0 1 m4 = x y z

1 0 1 1 m5 = x y z

1 1 0 0

1 1 1 0

F = ∑(0, 1, 2, 4, 5) =

m0 + m1 + m2 + m4 + m5 =

x y z + x y z + x y z +

x y z + x y z

Converting to Product-of-Maxterms Form

A function that is not in the Product-of-Minterms form can be converted to that form by means of a truth table

Consider again: F = y + x z

x y z F Minterm0 0 0 1

0 0 1 1

0 1 0 1

0 1 1 0 M3 = (x+y+z)

1 0 0 1

1 0 1 1

1 1 0 0 M6 = (x+y+z)

1 1 1 0 M7 = (x+y+z)

F = ∏(3, 6, 7) =

M3 · M6 · M7 =

(x+y+z) (x+y+z) (x+y+z)

Conversions Between Canonical Forms

F = m1+m2+m3+m5+m7 = ∑(1, 2, 3, 5, 7) =

x y z + x y z + x y z + x y z + x y z

F = M0 · M4 · M6 = ∏(0, 4, 6) = (x+y+z)(x+y+z)(x+y+z)

x y z F Minterm Maxterm0 0 0 0 M0 = (x + y + z)

0 0 1 1 m1 = x y z

0 1 0 1 m2 = x y z

0 1 1 1 m3 = x y z

1 0 0 0 M4 = (x + y + z)

1 0 1 1 m5 = x y z

1 1 0 0 M6 = (x + y + z)

1 1 1 1 m7 = x y z

Algebraic Conversion to Sum-of-Minterms

Expand all terms first to explicitly list all minterms AND any term missing a variable v with (v + v) Example 1: f = x + x y (2 variables)

f = x (y + y) + x yf = x y + x y + x yf = m3 + m2 + m0 = ∑(0, 2, 3)

Example 2: g = a + b c (3 variables)g = a (b + b)(c + c) + (a + a) b cg = a b c + a b c + a b c + a b c + a b c + a b cg = a b c + a b c + a b c + a b c + a b cg = m1 + m4 + m5 + m6 + m7 = ∑ (1, 4, 5, 6, 7)

Algebraic Conversion to Product-of-Maxterms

Expand all terms first to explicitly list all maxterms OR any term missing a variable v with v · v Example 1: f = x + x y (2 variables)

Apply 2nd distributive law:

f = (x + x) (x + y) = 1 · (x + y) = (x + y) = M1

Example 2: g = a c + b c + a b (3 variables)

g = (a c + b c + a) (a c + b c + b) (distributive)

g = (c + b c + a) (a c + c + b) (x + x y = x + y)

g = (c + b + a) (a + c + b) (x + x y = x + y)

g = (a + b + c) (a + b + c) = M5 . M2 = ∏ (2, 5)

Function Complements

The complement of a function expressed as a sum of minterms is constructed by selecting the minterms missing in the sum-of-minterms canonical form

Alternatively, the complement of a function expressed by a Sum of Minterms form is simply the Product of Maxterms with the same indices

Example: Given F(x, y, z) = ∑ (1, 3, 5, 7)

F(x, y, z) = ∑ (0, 2, 4, 6)

F(x, y, z) = ∏ (1, 3, 5, 7)

Summary of Minterms and Maxterms

There are 2n minterms and maxterms for Boolean functions with n variables.

Minterms and maxterms are indexed from 0 to 2n – 1

Any Boolean function can be expressed as a logical sum of minterms and as a logical product of maxterms

The complement of a function contains those minterms not included in the original function

The complement of a sum-of-minterms is a product-of-maxterms with the same indices

Standard Sum-of-Products (SOP) form: equations are written as an OR of AND terms

Standard Product-of-Sums (POS) form: equations are written as an AND of OR terms

Examples:• SOP:• POS:

These “mixed” forms are neither SOP nor POS• •

Standard Forms

B C B A C B A ++

C · )C B(A · B) (A +++

C) (A C) B (A ++

B) (A C A C B A ++

Standard Sum-of-Products (SOP)

A sum of minterms form for n variables can be written down directly from a truth table.• Implementation of this form is a two-level

network of gates such that:

• The first level consists of n-input AND gates

• The second level is a single OR gate

This form often can be simplified so that the corresponding circuit is simpler.

A Simplification Example: Writing the minterm expression:

F = A B C + A B C + A B C + ABC + ABC Simplifying:

F = A B C + A (B C + B C + B C + B C)F = A B C + A (B (C + C) + B (C + C))F = A B C + A (B + B)F = A B C + AF = B C + A

Simplified F contains 3 literals

Standard Sum-of-Products (SOP)

)7,6,5,4,1()C,B,A(F S=

AND/OR Two-Level Implementation

The two implementations for F are shown below

F

ABC

ABC

ABC

ABC

ABC

F

B

C

A

It is quite apparent which

is simpler!